We have seen in Chapter 4 that a spherical mirror gives spherical aberration, which we know from Section 1.7 to be independent of the location of its aperture stop. When the aperture stop is located at the center of curvature of the mirror, it also produces field curvature, although coma, astigmatism, and distortion are all zero. As we will discuss in Chapter 6, a paraboloidal mirror forms an aberration-free image of a point object only when it lies on its axis at an infinite distance from it. In order to utilize the simplicity of fabrication of a spherical mirror, we need a way to compensate its spherical aberration. An optical system consisting of a spherical mirror and a transparent plate of nonuniform thickness placed at its center of curvature to compensate for its spherical aberration is called a Schmidt camera. The plate is appropriately called the Schmidt plate. With the exception of field curvature, the image formed is free of primary aberrations.
As discussed in Section 4.4, the field curvature is such that an aberration-free image is formed on a spherical surface of radius of curvature equal to half that of the mirror. For an object at infinity, this surface is concentric with the mirror. In this chapter, we determine the shape of the Schmidt plate and discuss the chromatic aberrations associated with it. A numerical problem illustrates the results obtained.
5.2 SCHMIDT PLATE
Consider a spherical mirror with its aperture stop located at its center of curvature C, as shown in Figure 5-1, imaging an object lying at infinity. From Eq. (4-18), the optical path difference between a ray of zone r and the chief ray from an axial point object is given by
where f′ is the focal length of the mirror. It is negative, implying that the optical path length of the ray under consideration to the focus F′ is shorter than that of the chief ray. It also means that the ray intersects the axis after reflection at an axial point F″, which is slightly closer to the mirror vertex than the paraxial focus F′. This may be seen independently from the isosceles triangle CAF″ in which CF″ + AF″ > CA = 2 |f′| or CF″ > |f′| = CF′, since CF″ = AF″. In order that the path length of the ray be equal to that of the chief ray, its path length must be increased.